deploy

chapter_searching/binary_search/index.html

@@ -3418,7 +3418,7 @@
 <p>给定一个长度为 <span class="arithmatex">\(n\)</span> 的数组 <code>nums</code> ，元素按从小到大的顺序排列，数组不包含重复元素。请查找并返回元素 <code>target</code> 在该数组中的索引。若数组不包含该元素，则返回 <span class="arithmatex">\(-1\)</span> 。</p>
 </div>
 <p><img alt="二分查找示例数据" src="../binary_search.assets/binary_search_example.png" /></p>
-<p align="center"> Fig. 二分查找示例数据 </p>
+<p align="center"> 图：二分查找示例数据 </p>
 
 <p>对于上述问题，我们先初始化指针 <span class="arithmatex">\(i = 0\)</span> 和 <span class="arithmatex">\(j = n - 1\)</span> ，分别指向数组首元素和尾元素，代表搜索区间 <span class="arithmatex">\([0, n - 1]\)</span> 。请注意，中括号表示闭区间，其包含边界值本身。</p>
 <p>接下来，循环执行以下两个步骤：</p>
@@ -3457,6 +3457,8 @@
 </div>
 </div>
 </div>
+<p align="center"> 图：binary_search_step1 </p>
+
 <p>值得注意的是，由于 <span class="arithmatex">\(i\)</span> 和 <span class="arithmatex">\(j\)</span> 都是 <code>int</code> 类型，<strong>因此 <span class="arithmatex">\(i + j\)</span> 可能会超出 <code>int</code> 类型的取值范围</strong>。为了避免大数越界，我们通常采用公式 <span class="arithmatex">\(m = \lfloor {i + (j - i) / 2} \rfloor\)</span> 来计算中点。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="2:12"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><input id="__tabbed_2_6" name="__tabbed_2" type="radio" /><input id="__tabbed_2_7" name="__tabbed_2" type="radio" /><input id="__tabbed_2_8" name="__tabbed_2" type="radio" /><input id="__tabbed_2_9" name="__tabbed_2" type="radio" /><input id="__tabbed_2_10" name="__tabbed_2" type="radio" /><input id="__tabbed_2_11" name="__tabbed_2" type="radio" /><input id="__tabbed_2_12" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Java</label><label for="__tabbed_2_2">C++</label><label for="__tabbed_2_3">Python</label><label for="__tabbed_2_4">Go</label><label for="__tabbed_2_5">JS</label><label for="__tabbed_2_6">TS</label><label for="__tabbed_2_7">C</label><label for="__tabbed_2_8">C#</label><label for="__tabbed_2_9">Swift</label><label for="__tabbed_2_10">Zig</label><label for="__tabbed_2_11">Dart</label><label for="__tabbed_2_12">Rust</label></div>
 <div class="tabbed-content">
@@ -3988,7 +3990,7 @@
 <p>如下图所示，在两种区间表示下，二分查找算法的初始化、循环条件和缩小区间操作皆有所不同。</p>
 <p>在“双闭区间”表示法中，由于左右边界都被定义为闭区间，因此指针 <span class="arithmatex">\(i\)</span> 和 <span class="arithmatex">\(j\)</span> 缩小区间操作也是对称的。这样更不容易出错。因此，<strong>我们通常采用“双闭区间”的写法</strong>。</p>
 <p><img alt="两种区间定义" src="../binary_search.assets/binary_search_ranges.png" /></p>
-<p align="center"> Fig. 两种区间定义 </p>
+<p align="center"> 图：两种区间定义 </p>
 
 <h2 id="1012">10.1.2. &nbsp; 优点与局限性<a class="headerlink" href="#1012" title="Permanent link">&para;</a></h2>
 <p>二分查找在时间和空间方面都有较好的性能：</p>

chapter_searching/binary_search_edge/index.html

@@ -3589,7 +3589,7 @@
 <p>实际上，我们可以利用查找最左元素的函数来查找最右元素，具体方法为：<strong>将查找最右一个 <code>target</code> 转化为查找最左一个 <code>target + 1</code></strong>。</p>
 <p>查找完成后，指针 <span class="arithmatex">\(i\)</span> 指向最左一个 <code>target + 1</code>（如果存在），而 <span class="arithmatex">\(j\)</span> 指向最右一个 <code>target</code> ，<strong>因此返回 <span class="arithmatex">\(j\)</span> 即可</strong>。</p>
 <p><img alt="将查找右边界转化为查找左边界" src="../binary_search_edge.assets/binary_search_right_edge_by_left_edge.png" /></p>
-<p align="center"> Fig. 将查找右边界转化为查找左边界 </p>
+<p align="center"> 图：将查找右边界转化为查找左边界 </p>
 
 <p>请注意，返回的插入点是 <span class="arithmatex">\(i\)</span> ，因此需要将其减 <span class="arithmatex">\(1\)</span> ，从而获得 <span class="arithmatex">\(j\)</span> 。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="2:12"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><input id="__tabbed_2_6" name="__tabbed_2" type="radio" /><input id="__tabbed_2_7" name="__tabbed_2" type="radio" /><input id="__tabbed_2_8" name="__tabbed_2" type="radio" /><input id="__tabbed_2_9" name="__tabbed_2" type="radio" /><input id="__tabbed_2_10" name="__tabbed_2" type="radio" /><input id="__tabbed_2_11" name="__tabbed_2" type="radio" /><input id="__tabbed_2_12" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Java</label><label for="__tabbed_2_2">C++</label><label for="__tabbed_2_3">Python</label><label for="__tabbed_2_4">Go</label><label for="__tabbed_2_5">JS</label><label for="__tabbed_2_6">TS</label><label for="__tabbed_2_7">C</label><label for="__tabbed_2_8">C#</label><label for="__tabbed_2_9">Swift</label><label for="__tabbed_2_10">Zig</label><label for="__tabbed_2_11">Dart</label><label for="__tabbed_2_12">Rust</label></div>
@@ -3722,7 +3722,7 @@
 <li>查找最右一个 <code>target</code> ：可以转化为查找 <code>target + 0.5</code> ，并返回指针 <span class="arithmatex">\(j\)</span> 。</li>
 </ul>
 <p><img alt="将查找边界转化为查找元素" src="../binary_search_edge.assets/binary_search_edge_by_element.png" /></p>
-<p align="center"> Fig. 将查找边界转化为查找元素 </p>
+<p align="center"> 图：将查找边界转化为查找元素 </p>
 
 <p>代码在此省略，值得注意的有：</p>
 <ul>

chapter_searching/binary_search_insertion/index.html

@@ -3427,7 +3427,7 @@
 <p>给定一个长度为 <span class="arithmatex">\(n\)</span> 的有序数组 <code>nums</code> 和一个元素 <code>target</code> ，数组不存在重复元素。现将 <code>target</code> 插入到数组 <code>nums</code> 中，并保持其有序性。若数组中已存在元素 <code>target</code> ，则插入到其左方。请返回插入后 <code>target</code> 在数组中的索引。</p>
 </div>
 <p><img alt="二分查找插入点示例数据" src="../binary_search_insertion.assets/binary_search_insertion_example.png" /></p>
-<p align="center"> Fig. 二分查找插入点示例数据 </p>
+<p align="center"> 图：二分查找插入点示例数据 </p>
 
 <p>如果想要复用上节的二分查找代码，则需要回答以下两个问题。</p>
 <p><strong>问题一</strong>：当数组中包含 <code>target</code> 时，插入点的索引是否是该元素的索引？</p>
@@ -3586,7 +3586,7 @@
 <li>从索引 <span class="arithmatex">\(k\)</span> 开始，向左进行线性遍历，当找到最左边的 <code>target</code> 时返回。</li>
 </ol>
 <p><img alt="线性查找重复元素的插入点" src="../binary_search_insertion.assets/binary_search_insertion_naive.png" /></p>
-<p align="center"> Fig. 线性查找重复元素的插入点 </p>
+<p align="center"> 图：线性查找重复元素的插入点 </p>
 
 <p>此方法虽然可用，但其包含线性查找，因此时间复杂度为 <span class="arithmatex">\(O(n)\)</span> 。当数组中存在很多重复的 <code>target</code> 时，该方法效率很低。</p>
 <p>现考虑修改二分查找代码。整体流程不变，每轮先计算中点索引 <span class="arithmatex">\(m\)</span> ，再判断 <code>target</code> 和 <code>nums[m]</code> 大小关系：</p>
@@ -3623,6 +3623,8 @@
 </div>
 </div>
 </div>
+<p align="center"> 图：二分查找重复元素的插入点的步骤 </p>
+
 <p>观察以下代码，判断分支 <code>nums[m] &gt; target</code> 和 <code>nums[m] == target</code> 的操作相同，因此两者可以合并。</p>
 <p>即便如此，我们仍然可以将判断条件保持展开，因为其逻辑更加清晰、可读性更好。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="3:12"><input checked="checked" id="__tabbed_3_1" name="__tabbed_3" type="radio" /><input id="__tabbed_3_2" name="__tabbed_3" type="radio" /><input id="__tabbed_3_3" name="__tabbed_3" type="radio" /><input id="__tabbed_3_4" name="__tabbed_3" type="radio" /><input id="__tabbed_3_5" name="__tabbed_3" type="radio" /><input id="__tabbed_3_6" name="__tabbed_3" type="radio" /><input id="__tabbed_3_7" name="__tabbed_3" type="radio" /><input id="__tabbed_3_8" name="__tabbed_3" type="radio" /><input id="__tabbed_3_9" name="__tabbed_3" type="radio" /><input id="__tabbed_3_10" name="__tabbed_3" type="radio" /><input id="__tabbed_3_11" name="__tabbed_3" type="radio" /><input id="__tabbed_3_12" name="__tabbed_3" type="radio" /><div class="tabbed-labels"><label for="__tabbed_3_1">Java</label><label for="__tabbed_3_2">C++</label><label for="__tabbed_3_3">Python</label><label for="__tabbed_3_4">Go</label><label for="__tabbed_3_5">JS</label><label for="__tabbed_3_6">TS</label><label for="__tabbed_3_7">C</label><label for="__tabbed_3_8">C#</label><label for="__tabbed_3_9">Swift</label><label for="__tabbed_3_10">Zig</label><label for="__tabbed_3_11">Dart</label><label for="__tabbed_3_12">Rust</label></div>

chapter_searching/replace_linear_by_hashing/index.html

@@ -3420,7 +3420,7 @@
 <h2 id="1041">10.4.1. &nbsp; 线性查找：以时间换空间<a class="headerlink" href="#1041" title="Permanent link">&para;</a></h2>
 <p>考虑直接遍历所有可能的组合。开启一个两层循环，在每轮中判断两个整数的和是否为 <code>target</code> ，若是，则返回它们的索引。</p>
 <p><img alt="线性查找求解两数之和" src="../replace_linear_by_hashing.assets/two_sum_brute_force.png" /></p>
-<p align="center"> Fig. 线性查找求解两数之和 </p>
+<p align="center"> 图：线性查找求解两数之和 </p>
 
 <div class="tabbed-set tabbed-alternate" data-tabs="1:12"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><input id="__tabbed_1_7" name="__tabbed_1" type="radio" /><input id="__tabbed_1_8" name="__tabbed_1" type="radio" /><input id="__tabbed_1_9" name="__tabbed_1" type="radio" /><input id="__tabbed_1_10" name="__tabbed_1" type="radio" /><input id="__tabbed_1_11" name="__tabbed_1" type="radio" /><input id="__tabbed_1_12" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Java</label><label for="__tabbed_1_2">C++</label><label for="__tabbed_1_3">Python</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">JS</label><label for="__tabbed_1_6">TS</label><label for="__tabbed_1_7">C</label><label for="__tabbed_1_8">C#</label><label for="__tabbed_1_9">Swift</label><label for="__tabbed_1_10">Zig</label><label for="__tabbed_1_11">Dart</label><label for="__tabbed_1_12">Rust</label></div>
 <div class="tabbed-content">
@@ -3583,13 +3583,14 @@
 <div class="highlight"><span class="filename">two_sum.dart</span><pre><span></span><code><a id="__codelineno-10-1" name="__codelineno-10-1" href="#__codelineno-10-1"></a><span class="cm">/* 方法一： 暴力枚举 */</span>
 <a id="__codelineno-10-2" name="__codelineno-10-2" href="#__codelineno-10-2"></a><span class="n">List</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">twoSumBruteForce</span><span class="p">(</span><span class="n">List</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">nums</span><span class="p">,</span><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">target</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
 <a id="__codelineno-10-3" name="__codelineno-10-3" href="#__codelineno-10-3"></a><span class="w">  </span><span class="kt">int</span><span class="w"> </span><span class="n">size</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">nums</span><span class="p">.</span><span class="n">length</span><span class="p">;</span>
-<a id="__codelineno-10-4" name="__codelineno-10-4" href="#__codelineno-10-4"></a><span class="w">  </span><span class="k">for</span><span class="w"> </span><span class="p">(</span><span class="kd">var</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="m">0</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">size</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="m">1</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-10-5" name="__codelineno-10-5" href="#__codelineno-10-5"></a><span class="w">    </span><span class="k">for</span><span class="w"> </span><span class="p">(</span><span class="kd">var</span><span class="w"> </span><span class="n">j</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="m">1</span><span class="p">;</span><span class="w"> </span><span class="n">j</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">size</span><span class="p">;</span><span class="w"> </span><span class="n">j</span><span class="o">++</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-10-6" name="__codelineno-10-6" href="#__codelineno-10-6"></a><span class="w">      </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="n">nums</span><span class="p">[</span><span class="n">j</span><span class="p">]</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="n">target</span><span class="p">)</span><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="p">[</span><span class="n">i</span><span class="p">,</span><span class="w"> </span><span class="n">j</span><span class="p">];</span>
-<a id="__codelineno-10-7" name="__codelineno-10-7" href="#__codelineno-10-7"></a><span class="w">    </span><span class="p">}</span>
-<a id="__codelineno-10-8" name="__codelineno-10-8" href="#__codelineno-10-8"></a><span class="w">  </span><span class="p">}</span>
-<a id="__codelineno-10-9" name="__codelineno-10-9" href="#__codelineno-10-9"></a><span class="w">  </span><span class="k">return</span><span class="w"> </span><span class="p">[</span><span class="m">0</span><span class="p">];</span>
-<a id="__codelineno-10-10" name="__codelineno-10-10" href="#__codelineno-10-10"></a><span class="p">}</span>
+<a id="__codelineno-10-4" name="__codelineno-10-4" href="#__codelineno-10-4"></a><span class="w">  </span><span class="c1">// 两层循环，时间复杂度 O(n^2)</span>
+<a id="__codelineno-10-5" name="__codelineno-10-5" href="#__codelineno-10-5"></a><span class="w">  </span><span class="k">for</span><span class="w"> </span><span class="p">(</span><span class="kd">var</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="m">0</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">size</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="m">1</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
+<a id="__codelineno-10-6" name="__codelineno-10-6" href="#__codelineno-10-6"></a><span class="w">    </span><span class="k">for</span><span class="w"> </span><span class="p">(</span><span class="kd">var</span><span class="w"> </span><span class="n">j</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="m">1</span><span class="p">;</span><span class="w"> </span><span class="n">j</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">size</span><span class="p">;</span><span class="w"> </span><span class="n">j</span><span class="o">++</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
+<a id="__codelineno-10-7" name="__codelineno-10-7" href="#__codelineno-10-7"></a><span class="w">      </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="n">nums</span><span class="p">[</span><span class="n">j</span><span class="p">]</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="n">target</span><span class="p">)</span><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="p">[</span><span class="n">i</span><span class="p">,</span><span class="w"> </span><span class="n">j</span><span class="p">];</span>
+<a id="__codelineno-10-8" name="__codelineno-10-8" href="#__codelineno-10-8"></a><span class="w">    </span><span class="p">}</span>
+<a id="__codelineno-10-9" name="__codelineno-10-9" href="#__codelineno-10-9"></a><span class="w">  </span><span class="p">}</span>
+<a id="__codelineno-10-10" name="__codelineno-10-10" href="#__codelineno-10-10"></a><span class="w">  </span><span class="k">return</span><span class="w"> </span><span class="p">[</span><span class="m">0</span><span class="p">];</span>
+<a id="__codelineno-10-11" name="__codelineno-10-11" href="#__codelineno-10-11"></a><span class="p">}</span>
 </code></pre></div>
 </div>
 <div class="tabbed-block">
@@ -3630,6 +3631,8 @@
 </div>
 </div>
 </div>
+<p align="center"> 图：辅助哈希表求解两数之和 </p>
+
 <p>实现代码如下所示，仅需单层循环即可。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="3:12"><input checked="checked" id="__tabbed_3_1" name="__tabbed_3" type="radio" /><input id="__tabbed_3_2" name="__tabbed_3" type="radio" /><input id="__tabbed_3_3" name="__tabbed_3" type="radio" /><input id="__tabbed_3_4" name="__tabbed_3" type="radio" /><input id="__tabbed_3_5" name="__tabbed_3" type="radio" /><input id="__tabbed_3_6" name="__tabbed_3" type="radio" /><input id="__tabbed_3_7" name="__tabbed_3" type="radio" /><input id="__tabbed_3_8" name="__tabbed_3" type="radio" /><input id="__tabbed_3_9" name="__tabbed_3" type="radio" /><input id="__tabbed_3_10" name="__tabbed_3" type="radio" /><input id="__tabbed_3_11" name="__tabbed_3" type="radio" /><input id="__tabbed_3_12" name="__tabbed_3" type="radio" /><div class="tabbed-labels"><label for="__tabbed_3_1">Java</label><label for="__tabbed_3_2">C++</label><label for="__tabbed_3_3">Python</label><label for="__tabbed_3_4">Go</label><label for="__tabbed_3_5">JS</label><label for="__tabbed_3_6">TS</label><label for="__tabbed_3_7">C</label><label for="__tabbed_3_8">C#</label><label for="__tabbed_3_9">Swift</label><label for="__tabbed_3_10">Zig</label><label for="__tabbed_3_11">Dart</label><label for="__tabbed_3_12">Rust</label></div>
 <div class="tabbed-content">
@@ -3834,15 +3837,17 @@
 <div class="highlight"><span class="filename">two_sum.dart</span><pre><span></span><code><a id="__codelineno-22-1" name="__codelineno-22-1" href="#__codelineno-22-1"></a><span class="cm">/* 方法二： 辅助哈希表 */</span>
 <a id="__codelineno-22-2" name="__codelineno-22-2" href="#__codelineno-22-2"></a><span class="n">List</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">twoSumHashTable</span><span class="p">(</span><span class="n">List</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">nums</span><span class="p">,</span><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="n">target</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
 <a id="__codelineno-22-3" name="__codelineno-22-3" href="#__codelineno-22-3"></a><span class="w">  </span><span class="kt">int</span><span class="w"> </span><span class="n">size</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">nums</span><span class="p">.</span><span class="n">length</span><span class="p">;</span>
-<a id="__codelineno-22-4" name="__codelineno-22-4" href="#__codelineno-22-4"></a><span class="w">  </span><span class="n">Map</span><span class="o">&lt;</span><span class="kt">int</span><span class="p">,</span><span class="w"> </span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">dic</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">HashMap</span><span class="p">();</span>
-<a id="__codelineno-22-5" name="__codelineno-22-5" href="#__codelineno-22-5"></a><span class="w">  </span><span class="k">for</span><span class="w"> </span><span class="p">(</span><span class="kd">var</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="m">0</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">size</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-22-6" name="__codelineno-22-6" href="#__codelineno-22-6"></a><span class="w">    </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">dic</span><span class="p">.</span><span class="n">containsKey</span><span class="p">(</span><span class="n">target</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]))</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-22-7" name="__codelineno-22-7" href="#__codelineno-22-7"></a><span class="w">      </span><span class="k">return</span><span class="w"> </span><span class="p">[</span><span class="n">dic</span><span class="p">[</span><span class="n">target</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]]</span><span class="o">!</span><span class="p">,</span><span class="w"> </span><span class="n">i</span><span class="p">];</span>
-<a id="__codelineno-22-8" name="__codelineno-22-8" href="#__codelineno-22-8"></a><span class="w">    </span><span class="p">}</span>
-<a id="__codelineno-22-9" name="__codelineno-22-9" href="#__codelineno-22-9"></a><span class="w">    </span><span class="n">dic</span><span class="p">.</span><span class="n">putIfAbsent</span><span class="p">(</span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">],</span><span class="w"> </span><span class="p">()</span><span class="w"> </span><span class="o">=&gt;</span><span class="w"> </span><span class="n">i</span><span class="p">);</span>
-<a id="__codelineno-22-10" name="__codelineno-22-10" href="#__codelineno-22-10"></a><span class="w">  </span><span class="p">}</span>
-<a id="__codelineno-22-11" name="__codelineno-22-11" href="#__codelineno-22-11"></a><span class="w">  </span><span class="k">return</span><span class="w"> </span><span class="p">[</span><span class="m">0</span><span class="p">];</span>
-<a id="__codelineno-22-12" name="__codelineno-22-12" href="#__codelineno-22-12"></a><span class="p">}</span>
+<a id="__codelineno-22-4" name="__codelineno-22-4" href="#__codelineno-22-4"></a><span class="w">  </span><span class="c1">// 辅助哈希表，空间复杂度 O(n)</span>
+<a id="__codelineno-22-5" name="__codelineno-22-5" href="#__codelineno-22-5"></a><span class="w">  </span><span class="n">Map</span><span class="o">&lt;</span><span class="kt">int</span><span class="p">,</span><span class="w"> </span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">dic</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">HashMap</span><span class="p">();</span>
+<a id="__codelineno-22-6" name="__codelineno-22-6" href="#__codelineno-22-6"></a><span class="w">  </span><span class="c1">// 单层循环，时间复杂度 O(n)</span>
+<a id="__codelineno-22-7" name="__codelineno-22-7" href="#__codelineno-22-7"></a><span class="w">  </span><span class="k">for</span><span class="w"> </span><span class="p">(</span><span class="kd">var</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="m">0</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">&lt;</span><span class="w"> </span><span class="n">size</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
+<a id="__codelineno-22-8" name="__codelineno-22-8" href="#__codelineno-22-8"></a><span class="w">    </span><span class="k">if</span><span class="w"> </span><span class="p">(</span><span class="n">dic</span><span class="p">.</span><span class="n">containsKey</span><span class="p">(</span><span class="n">target</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]))</span><span class="w"> </span><span class="p">{</span>
+<a id="__codelineno-22-9" name="__codelineno-22-9" href="#__codelineno-22-9"></a><span class="w">      </span><span class="k">return</span><span class="w"> </span><span class="p">[</span><span class="n">dic</span><span class="p">[</span><span class="n">target</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]]</span><span class="o">!</span><span class="p">,</span><span class="w"> </span><span class="n">i</span><span class="p">];</span>
+<a id="__codelineno-22-10" name="__codelineno-22-10" href="#__codelineno-22-10"></a><span class="w">    </span><span class="p">}</span>
+<a id="__codelineno-22-11" name="__codelineno-22-11" href="#__codelineno-22-11"></a><span class="w">    </span><span class="n">dic</span><span class="p">.</span><span class="n">putIfAbsent</span><span class="p">(</span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">],</span><span class="w"> </span><span class="p">()</span><span class="w"> </span><span class="o">=&gt;</span><span class="w"> </span><span class="n">i</span><span class="p">);</span>
+<a id="__codelineno-22-12" name="__codelineno-22-12" href="#__codelineno-22-12"></a><span class="w">  </span><span class="p">}</span>
+<a id="__codelineno-22-13" name="__codelineno-22-13" href="#__codelineno-22-13"></a><span class="w">  </span><span class="k">return</span><span class="w"> </span><span class="p">[</span><span class="m">0</span><span class="p">];</span>
+<a id="__codelineno-22-14" name="__codelineno-22-14" href="#__codelineno-22-14"></a><span class="p">}</span>
 </code></pre></div>
 </div>
 <div class="tabbed-block">

chapter_searching/searching_algorithm_revisited/index.html

@@ -3457,7 +3457,7 @@
 <h2 id="1053">10.5.3. &nbsp; 搜索方法选取<a class="headerlink" href="#1053" title="Permanent link">&para;</a></h2>
 <p>给定大小为 <span class="arithmatex">\(n\)</span> 的一组数据，我们可以使用线性搜索、二分查找、树查找、哈希查找等多种方法在该数据中搜索目标元素。各个方法的工作原理如下图所示。</p>
 <p><img alt="多种搜索策略" src="../searching_algorithm_revisited.assets/searching_algorithms.png" /></p>
-<p align="center"> Fig. 多种搜索策略 </p>
+<p align="center"> 图：多种搜索策略 </p>
 
 <p>上述几种方法的操作效率与特性如下表所示。</p>
 <div class="center-table">